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Charged dielectric sheet between two grounded conductors

  1. Aug 31, 2011 #1
    I have a proposed solution to the following problem, but I don't know if is correct, and so I'll ask you for help:

    A thin dielectric sheet of uniform charge density [itex]\sigma[/itex] (I'll assume [itex]\sigma > 0[/itex] for simplicity) is surrounded by two thin conductors, which are initially uncharged and are connected by a wire. The first and second conductors are at distances [itex]a, b[/itex] respectively from the dielectric. The three sheets are parallel. The configuration is then like a sandwich. I want to find the electric field in every point of the space.

    My proposed solution is that the electric field between the conductor A and the dielectric, and between the conductor B and the dielectric, is null. The electric field is [itex]|\bar{E}_{outside}| = \dfrac{\sigma}{2\epsilon_0}[/itex] everywhere else.

    My argumentation is as follows: for the electric field outside the sandwich, the solution is clear and comes from Gauss's law.

    For the electric field inside, let's suppose that a charge [itex]\eta[/itex] is induced on the sheet A (and thus a charge [itex]-\eta[/itex] on B). This means that the electric field in the space between A and the dielectric will be [itex]E_{A} = \dfrac{\sigma+\eta}{2\epsilon_0}[/itex], while [itex]E_{B} = \dfrac{\sigma-\eta}{2\epsilon_0}[/itex]. The potential difference between A and B is then [itex]\Delta V = E_{B} b - E_{A} a[/itex].

    But the hypothesis is that [itex]\Delta V = 0[/itex] (they are connected with a wire), and so the difference above only holds is [itex]\eta = \sigma[/itex], which means that [itex]E_{A} = E_{B} = 0[/itex].
     
  2. jcsd
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